(a) Two vertices of a triangle are (5,...

MathematicsCentroid, Incenter, Orthocenter, and CircumcenterJEE Advanced 1979Moderate

(a) Two vertices of a triangle are $(5, - 1)$ and $(- 2, 3)$ . If the orthocentre of the triangle is the origin, find the coordinates of the third point. (b) Find the equation of the line which bisects the obtuse angle between the lines $x - 2 y + 4 = 0$ and $4 x - 3 y + 2 = 0$ .

Answer:(a) (-4, -7), (b) (4-\sqrt{5})x + (2\sqrt{5}-3)y - (4\sqrt{5}-2) = 0

Visualized Solution (Hindi)

Visualizing the Triangle and Orthocenter

Given vertices: $A (5, - 1)$ and $B (- 2, 3)$
Orthocenter: $H (0, 0)$
Let the third vertex be $C (h, k)$

Property of Orthocenter: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A H ⊥ B C

Property: The altitude from $A$ is perpendicular to $B C$ .
Therefore, Slope of $A H \times$ Slope of $B C = - 1$

Setting up the First Equation

Slope of $A H = \frac{- 1 - 0}{5 - 0} = - \frac{1}{5}$
Slope of $B C = \frac{k - 3}{h - ( - 2 )} = \frac{k - 3}{h + 2}$
Equation: $(- \frac{1}{5}) \times (\frac{k - 3}{h + 2}) = - 1$

Simplifying to Linear Form

$k - 3 = 5 (h + 2)$
$k - 3 = 5 h + 10$
Equation 1: $5 h - k + 13 = 0$

Property of Orthocenter: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B H ⊥ A C

Property: The altitude from $B$ is perpendicular to $A C$ .
Therefore, Slope of $B H \times$ Slope of $A C = - 1$

Setting up the Second Equation

Slope of $B H = \frac{3 - 0}{- 2 - 0} = - \frac{3}{2}$
Slope of $A C = \frac{k - ( - 1 )}{h - 5} = \frac{k + 1}{h - 5}$
Equation: $(- \frac{3}{2}) \times (\frac{k + 1}{h - 5}) = - 1$

Simplifying the Second Equation

$3 (k + 1) = 2 (h - 5)$
$3 k + 3 = 2 h - 10$
Equation 2: $2 h - 3 k - 13 = 0$

Solving for Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C (h, k)

From Eq 1: $k = 5 h + 13$
Substitute in Eq 2: $2 h - 3 (5 h + 13) - 13 = 0$
$- 13 h - 39 - 13 = 0 \Rightarrow - 13 h = 52 \Rightarrow h = - 4$
$k = 5 (- 4) + 13 = - 7$
Vertex $C$ is $(- 4, - 7)$

Part (b): Lines and Angle Bisectors

Line 1: $x - 2 y + 4 = 0$
Line 2: $4 x - 3 y + 2 = 0$
Goal: Find the obtuse angle bisector.

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{1} a_{2} + b_{1} b_{2} Test

Ensure $c_{1}, c_{2} > 0$ : $4 > 0$ and $2 > 0$ .
Calculate $a_{1} a_{2} + b_{1} b_{2} = (1) (4) + (- 2) (- 3) = 10$
Since $a_{1} a_{2} + b_{1} b_{2} > 0$ , the positive sign gives the obtuse bisector.

Applying the Bisector Formula

Formula: $\frac{a _{1} x + b _{1} y + c _{1}}{a _{1}^{2} + b _{1}^{2}} = + \frac{a _{2} x + b _{2} y + c _{2}}{a _{2}^{2} + b _{2}^{2}}$
$\frac{x - 2 y + 4}{1 ^{2} + ( - 2 ) ^{2}} = \frac{4 x - 3 y + 2}{4 ^{2} + ( - 3 ) ^{2}}$
$\frac{x - 2 y + 4}{5} = \frac{4 x - 3 y + 2}{5}$

Final Equation of the Bisector

$5 (x - 2 y + 4) = 5 (4 x - 3 y + 2)$
Divide by $5$ : $5 (x - 2 y + 4) = 4 x - 3 y + 2$
$(4 - 5) x + (25 - 3) y - (45 - 2) = 0$

Summary and Key Takeaways

Key Takeaway 1: Orthocenter $H$ means $A H ⊥ B C$ and $B H ⊥ A C$ .
Key Takeaway 2: For obtuse bisector, check sign of $a_{1} a_{2} + b_{1} b_{2}$ after making $c_{1}, c_{2} > 0$ .
Final Answer (a): $(- 4, - 7)$
Final Answer (b): $(4 - 5) x + (25 - 3) y - (45 - 2) = 0$

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